# Finite population correction factor

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Whenever we take a random sample from a population having variance σ, we know that the sample mean has variance given as σ/√n where n is the sample size. But this is only true if we assume that:

1.  The sample has been chosen with replacement.
2.  The population size is infinite.

Both of these assumptions are false in real-life situations. But the above estimate for the variance of the sampling distribution still works as an approximation as long as the sample size is less than 5% of the total population.

If the sample size exceeds 5% of the population then we need to multiply the standard deviation of the sample mean by a correction factor called the FPC (Finite population correction factor).

### Formula for the finite population correction factor:

The FPC can be calculated using the formula,

Thus the formula for standard error of sampling mean becomes,

And the formula for standard error of sampling proportion becomes,

Example 1: Suppose that a population of size 10000 has a variance of 100. We want to estimate the average height of people in the population. If a sample of size 2500 is selected from the population then calculate the standard error of the sample mean.

Solution: The formula for standard error is given as S.E = σ/√n but since the sample is of size greater than 5% of the population (=500) therefore we must apply the correction factor to the standard error.

The formula becomes, S.E = √[(N-n)/(N-1)]* σ /√n = √(7500/9999) * 10/50 = 0.17321

Example 2: Suppose that we have a population of size 1000 that has a variance of 25. We obtain an estimate for the proportion ‘p’ of blue-eyed people in the population as p=0.5. If we have drawn a sample of size 100 from the population then calculate the standard error of the population proportion

Solution: The formula for standard error is given as, S.E = √p(1-p)/n but since the sample is of size greater than 5% of the population (=50) therefore we must apply the correction factor to the standard error.

The formula becomes, S.E = √[(N-n)/(N-1)]* √p(1-p)/n = √(900/999) * √(0.5*0.5/100) = 0.0474

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